search for: 0.047

On Mon, Apr 13, 2009 at 12:41 PM, Dan Dube <ddube-at-advisen.com> wrote: > i use tapply and by often, but i always end up banging my head against > the wall with the output. The proposed solution of Dan's problem posted on R-help was: > do.call(rbind,a) When I use this 'solution' I get 'ERROR: second argument must be a list'. So head on wall continues. My

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

Hello, I’m trying to follow the syntax of a script from a journal website. In order to create a regression formula used later in the script, the regression matrix must have column names “X1”, “X2”, etc. I have tried to assign these column names to my matrix ScoutRSM.mat using a for loop, but I don’t know how to interpret the error message. Suggestions? Thanks, Paul

I have a dataset X Y1 1200 1.375 4000 0.464 1333.33 0.148 444.44 0.047 148.148 0.014 49.383 0.005 16.461 0.004 I have to find a curve fit for the above dataset based on a 4-parameter logistic equation viz. Y1 = d + ((a-d)/(1+(X/cc)^b)), where X and Y1 are the values above. I need to know how to solve the above equation for values a, b, c, d. -- View this message in context:

My problem is there I download and archive my work to the freebsd server via samba. When I'm transfering files from the windows -> freebsd I will get anywhere between 20-100kB/s and from freebsd -> windows I will get a few mbps. I'm getting no where near a full 100mbps and both ethernet cards are set for 100mbps full duplex working great. I've tried increasing buffer sizes on

Hi all I am trying to generate a normal unbalanced data to estimate the coefficients of LM, LMM, GLM, and GLMM and their standard errors. Also, I am trying to estimate the variance components and their standard errors. Further, I am trying to use the likelihood ratio test to test H0: sigma^2_b = 0 (random effects variance component), and the t-test to test H0:mu=0 (intercept of the model Yij = mu

Hi R users: What is the difference between in the computation of the diag of the "hat" matrix in: "lm.influence" and the matrix operations with "solve()" and "t()"? I mean, this is my X matrix x1 x2 x3 x4 x5 [1,] 0.297 0.310 0.290 0.220 0.1560 [2,] 0.360 0.390 0.369 0.297 0.2050 [3,] 0.075 0.058 0.047 0.034 0.0230 [4,] 0.114 0.100

I am trying to use t.test on the following data: date type INTERVAL nCASES MTF SDF MTO SDO nFST MF nOBS MO MB BIASCV BIASEV ME MAE RMSE CRCF 2001-06-15 avn GE1.00 4385 0.246 0.300 1.502 0.556 1367 1.373 4385 1.502 1.471 0.285 0.164 -1.256 1.266 1.399 0.056 2001-06-15 avn

Hi, I have fit a series of ols() models, by group, in this manner: l <- ols(y ~ rcs(x, 4)) ... where the series of 'x' values in each group is the same, however knots are not always identical between groups. The result is a table of 'coefs' derived from the ols objects, by group: group Intercept top top' top'' 1 6.864 0.01 2.241 -2.65

I can't get my simultaneous equations to work using system fit. Please help. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) str(Empdata) Empdata$gnipc<-as.numeric(Empdata$gnipc) install.packages("systemfit") library("systemfit") pdata <- plm.data(Empdata,

This concerns whether p-values from lmer can be trusted. From simulations, it seems that lmer can produce very small, and probably spurious, p-values. I realize that lmer is not yet a finished product. Is it likely that the problem will be fixed in a future release of the lme4 package? Using simulated data for a quite standard mixed-model anova (a balanced two-way design; see code for the

I have a data frame with 1 factor, one date, and 37 numeric values: str(waterchem) 'data.frame': 3525 obs. of 39 variables: site : Factor w/ 64 levels "D-1","D-2","D-3",..: 1 1 1 1 1 ... $ sampdate : Date, format: "2007-12-12" "2008-03-15" ... $ CO3 : num 1 1 6.7 1 1 1 1 1 1 1 ... $ HCO3 : num 231 228 118 246

OK, Let's try this again! Here is the reproducible script; it is long because I had to copy the panel dataset here. My question is related to systemfit; I don't know how to get the result for the entire panel. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) install.packages("systemfit")

Sadly you failed to set your email program to send plain text and the data is corrupted at my end. I also think you need to reduce the size of the data set... the intent here is to increase your understanding, not debug your particular analysis. I will say that I am having a very challenging time understanding what you are trying to accomplish though. What are the equations that you think need

Hi List, I don't understand why 'lrm' doesn't recognize the '~.' formula. I'm pretty sure it was working before. Please see below: I'm using R2.3.0, WinXP, Design 2.0-12 thanks, ...Tao > dat <- data.frame(y=factor(rep(1:2,each=50)), x1=rnorm(100), x2=rnorm(100), x3=rnorm(100)) > lrm(y~., data=dat, x=T, y=T) Error in terms.formula(formula, specials =

Hej! I have the following problem: I would like to do partial correlations on non-parametric data. I checked "pcor" (Computes the partial correlation between two variables given a set of other variables) but I do not know how to change to a Spearman Rank Correlation method [pcor(c("BCDNA","ImProd","A365"),var(PCor))] Here''s a glimpse of

Dear list, I have a data frame where one of the columns are p values with scientific notation mixed with regular numbers with decimals. >a=data frame >a P OR N 0.50 0.7500 237 0.047 1.1030 237 0.124 0.7742 237 0.124 0.7742 237 0.0080 1.1590 237 0.50 0.7500 237 4.5e-07 1.2 237 5.6e-04 0.9 237 when I try to do >pval=a$P/2 R gives me an error saying "In Ops.factor(pval, 2)

Hi, i'm newbie for this, but it's very interesting, but how i have to interpret the results if i get i.e. this results ? Is it correct - if the "Ratio of effect sd" is positiv than the Numerator effects are bigger , and the negative case vice-versa ? Ratio of effect standard deviations: 0.954 Log(sd ratio): -0.047 (se 0.828) Approximate 95% confidence

Hello, I've noticed that dget() is much slower in the current and devel R versions than in previous versions. In 2.15 reading a 10000-row data.frame takes less than half a second: > (which.r <- R.Version()$version.string) [1] "R version 2.15.2 (2012-10-26)" > x <- data.frame(matrix(sample(letters, 100000, replace = TRUE), ncol = 10)) > dput(x, which.r) >

Hello, I'm using the "sem" package to do a confirmatory factor analysis on data collected with a questionnaire. In the model, there is a unique factor G and 23 items. I would like to calculate the standardized residual variance of the observed variables. "Sem" only gives the residual variance with the "summary" function, or the standardized loadings with the